a

b

c

enclosedqSdE

0

Fundamental Lawof Electrostatics

• Coulomb’s LawForce between two point charges

OR

• Gauss’ LawRelationship between Electric Fields and charges

dS dS

1 2

2) A positive charge is contained inside a spherical shell. How does the electric flux dФE through the surface element dS change when the charge is moved from position 1 to position 2?

a) dФE increases

b) dФE decreases

c) dФE doesn’t change

Preflight 4:

dS dS

1 2

3) A positive charge is contained inside a spherical shell. How does the flux ФE through the entire surface change when the charge is moved from position 1 to position 2?

a) ФE increases

b) ФE decreases

c) ФE doesn’t change

Gauss’ Law• Gauss’ Law (a FUNDAMENTAL LAW):

The net electric flux through any closed surface is proportional to the charge enclosed by that surface.

• How do we use this equation??• The above equation is ALWAYS TRUE but it doesn’t

look easy to use.• It is very useful in finding E when the physical

situation exhibits massive SYMMETRY.

0

enclosedE

qE dS

™

Gauss’ Law…made easy

• To solve the above equation for E, you have to be able to CHOOSE A CLOSED SURFACE such that the integral is TRIVIAL.

(1) Direction: surface must be chosen such that E is known to be either parallel or perpendicular to each piece of the surface;

If then

If then

(2) Magnitude: surface must be chosen such that E has the same value at all points on the surface when E is perpendicular to the surface.

Sd||E

SdE

0SdE

E dS E dS

0/E enclosedE dS q

˜

• With these two conditions we can bring E outside of the integral…and:

Note that is just the area of the Gaussian surface over which we are integrating. Gauss’ Law now takes the form:

This equation can now be solved for E (at the surface) if we know qenclosed (or for qenclosed if we know E).

Gauss’ Law…made easy

dSEEdSSdE

dS

0

enclosedqdSE

0/E enclosedE dS q

˜

Geometry and Surface Integrals• If E is constant over a surface, and normal to it everywhere, we

can take E outside the integral, leaving only a surface area

z

R

RLRdS 22 2 L24 RdS R

dSESdE

you may use different E’sfor different surfaces

of your “object”

a b

c

x

y

z

abbcacdS 222

Gauss Þ Coulomb• We now illustrate this for the field of the

point charge and prove that Gauss’ Law implies Coulomb’s Law.

• Symmetry Þ E-field of point charge is radial and spherically symmetric

• Draw a sphere of radius R centered on the charge.

E

+QR

204

1RQE

•Why?E normal to every point on the surface

E has same value at every point on the surface can take E outside of the integral!

•Therefore, !

–Gauss’ Law

–We are free to choose the surface in such problems… we call this a “Gaussian”

surface

EdSSdE Þ

ERdSEEdSSdE 24

04R2E Q

Þ

Uniform charged sphere

• Outside sphere: (r>a)– We have spherical symmetry centered on the center of

the sphere of charge– Therefore, choose Gaussian surface = hollow sphere of

radius r

What is the magnitude of the electric field due to a solid sphere of radius a with uniform charge density r (C/m3)?

ar

r

0

24

qErSdE

r 3

34 aq

ÞGauss’

Law2

0

3

3 raE

r

20

14

qr

same as point charge!

Uniform charged sphere• Outside sphere: (r > a)

• Inside sphere: (r < a)– We still have spherical symmetry centered on the center of

the sphere of charge.– Therefore, choose Gaussian surface = sphere of radius r

20

3

3 raE

r

ar

r

Gauss’ Law

0

24

qErSdE

r 3r34qBut,

rE03

r

Thus:

ra

E

Gauss’ Law and Conductors• We know that E=0 inside a conductor (otherwise

the charges would move).

• But since .0E dS

inside 0Q

Charges on a conductor only reside on the surface(s)!

Conducting sphere

+++

++

++

+

1

A B

A blue sphere A is contained within a red spherical shell B. There is a charge QA on the blue sphere and charge QB on the red spherical shell.

7) The electric field in the region between the spheres is completely independent of QB the charge on the red spherical shell.

True

False

Preflight 4:

Consider the following two topologies:A) A solid non-conducting sphere

carries a total charge Q = -3 mC distributed evenly throughout. It is surrounded by an uncharged conducting spherical shell.

1A • Compare the electric field at point X in cases A and B:

(a) EA < EB (b) EA = EB (c) EA > EB

E

s2

s1

-|Q|

1B • What is the surface charge density s1 on the inner surface of the conducting shell in case A?

(a) s1 < 0 (b) s1 = 0 (c) s1 > 0

B) Same as (A) but conducting shell removed

Consider the following two topologies:

A) A solid non-conducting sphere carries a total charge Q = -3 mC distributed evenly throughout. It is surrounded by an uncharged conducting spherical shell. E

s2

s1

-|Q|

1A • Compare the electric field at point X in cases A and B:

(a) EA < EB (b) EA = EB (c) EA > EB

• Select a sphere passing through the point X as the Gaussian surface.• How much charge does it enclose?

• Answer: -|Q|, whether or not the uncharged shell is present.

(The field at point X is determined only by the objects with NET CHARGE.)

Consider the following two topologies:A solid non-conducting sphere carries a total charge Q = -3 mC and is surrounded by an uncharged conducting spherical shell.

B) Same as (A) but conducting shell removed

1B• What is the surface charge density s1 on the inner

surface of the conducting shell in case A?

(a) s1 < 0 (b) s1 = 0 (c) s1 > 0

E

s2

s1

• Inside the conductor, we know the field E = 0• Select a Gaussian surface inside the conductor

• Since E = 0 on this surface, the total enclosed charge must be 0• Therefore, s1 must be positive, to cancel the charge -|Q|

• By the way, to calculate the actual value: s1 = -Q / (4 r12)

-|Q|

Infinite Line of Charge• Symmetry Þ E-field

must be to line and can only depend on distance from line

• Therefore, CHOOSE Gaussian surface to be a cylinder of radius r and length h aligned with the x-axis.

• Apply Gauss’ Law:0 SdE

• On the ends,

rhESdE 2

hq AND• On the barrel, Þ rE

02

NOTE: we have obtained here the same result as we did last lecture using Coulomb’s Law. The symmetry makes today’s derivation easier.

+ + + + ++ + + +x

y

+ + + + + + + + + + +

Er

h

Er

+ + + + + ++ + +

2

• A line charge (C/m) is placed along the axis of an uncharged conducting cylinder of inner radius ri = a, and outer radius ro = b as shown. – What is the value of the charge density so

(C/m2) on the outer surface of the cylinder?

(a) (b) (c)bo

s2

0os

a

b

s0?

bo s

2

• A line charge (C/m) is placed along the axis of an uncharged conducting cylinder of inner radius ri = a, and outer radius ro = b as shown. – What is the value of the charge density so

(C/m2) on the outer surface of the cylinder?

(a) (b) (c)bo

s2

0os

a

b

s0?

bo s

2

View end on:

b

rEoutside

02

Draw Gaussian tube which surrounds only the outer edge

00

2)2()2(s

bLqErLErLEdS o

outsideconductor

so

brb

rE o

ooutside

ss

22 00

Þ

0

5) Given an infinite sheet of charge as shown in the figure. You need to use Gauss' Law to calculate the electric field near the sheet of charge. Which of the following Gaussian surfaces are best suited for this purpose?

a) a cylinder with its axis along the planeb) a cylinder with its axis perpendicular to the planec) a cubed) a sphere

Note: you may choose more than one answer

Preflight 4:

Infinite sheet of charge, surface charge density s

• Symmetry:direction of E = x-axis

• Therefore, CHOOSE Gaussian surface to be a cylinder whose axis is aligned with the x-axis. x

s

A

E E

02s

E

Conclusion: An infinite plane sheet of charge creates a CONSTANT electric field .

• Apply Gauss' Law:0 SdE

AESdE 2

• The charge enclosed = s A AEA20 s Therefore, Gauss’ Law Þ

• On the barrel,• On the ends,

Two Infinite Sheets(into the screen)

• Field outside must be zero. Two ways to see:

– Superposition

– Gaussian surface encloses zero charge

E=0 E=0s

+

+

++

++

+

s

+

++

-

-

--

-

--

-

-

-+

+

---

A

E

A

0AQ sinsideoutside AEAESdE

0s

E

• Field inside is NOT zero:– Superposition

– Gaussian surface encloses non-zero charge

(See also: mandatory on-line “tutorial exercises”)

enclosedqSdE

0• Gauss’ Law is ALWAYS VALID!

Gauss’ Law: Help for the Homework Problems

• What Can You Do With This?If you have (a) spherical, (b) cylindrical, or (c) planar symmetry AND:

• If you know the charge (RHS), you can calculate the electric field (LHS)• If you know the field (LHS, usually because E=0 inside conductor), you

can calculate the charge (RHS).

ErSdE 200 4

LHS:RHS: q = ALL charge inside radius r 2

041

rqE

rLESdE 200

LHS:RHS: q = ALL charge inside radius r, length L r

E02

AESdE 200

LHS:RHS: q = ALL charge inside cylinder =s A 02

sE

• Spherical Symmetry: Gaussian surface = sphere of radius r

• Planar Symmetry: Gaussian surface = cylinder of area A

• Cylindrical symmetry: Gaussian surface = cylinder of radius r

Sheets of Chargeσ1 σRσL

A B C D

Uncharged Conductor

Hints:1. Assume σ is positive. If it’s negative, the answer will still work.2. Assume to the right.3. Use superposition, but keep signs straight4. Think about which way a test charge would move.

ˆ+x

1 1 1A B C D

0 0 0

σ +σ σˆ ˆ ˆE x E x E 0 E x2E 2E 2E

E1 + EL - ER = 0σ1 + σ L - σ R = 0 σ L + σ R = 0 (uncharged conductor)

σ1 + 2σ L = 0

σ L = σ R = 1σ

2 1+σ

2

Summary• Gauss’ Law: Electric field flux through a

closed surface is proportional to the net charge enclosed– Gauss’ Law is exact and always true….

• Gauss’ Law makes solving for E-field easy when the symmetry is sufficient– spherical, cylindrical, planar

• Gauss’ Law proves that electric fields vanish in conductor– extra charges reside on surface

enclosed00 qSdE

Example 1: spheres• A solid conducting sphere is concentric

with a thin conducting shell, as shown• The inner sphere carries a charge Q1, and

the spherical shell carries a charge Q2, such that Q2 = -3Q1.

• How is the charge distributed on the sphere?

• How is the charge distributed on the spherical shell?

• What is the electric field at r < R1? Between R1 and R2? At r > R2?

• What happens when you connect the two spheres with a wire? (What are the

charges?)

R1

R2

Q1

Q2

A

B

C

D

• How is the charge distributed on the sphere?A

* The electric field inside a conductor is zero.

(A) By Gauss’s Law, there can be no net charge inside the conductor, and the charge must reside on the outside surface of the sphere

+++

+ ++

++

R1

R2

Q1

Q2

• How is the charge distributed on the spherical shell?B

* The electric field inside the conducting shell is zero.

(B) There can be no net charge inside the conductor, therefore the inner surface of the shell must carry a net charge of -Q1, and the outer surface must carry the charge +Q1 + Q2, so that the net charge on the shell equals Q2.

The charges are distributed uniformly over the inner and outer surfaces of the shell, hence

22

1inner R4

Q

s and2

2

12

2

12outer R4

Q2R4

QQ

s

R1

R2

Q1

Q2

(C) r < R1: Inside the conducting sphere

* The electric field inside a conductor is zero.

.0E

(C) Between R1 and R2 : R1 < r < R2

Charge enclosed = Q1

1

2ˆQE k r

r

(C) r > R2

Charge enclosed = Q1 + Q2

1 2 1

2 20

1 2ˆ ˆ4

Q Q QE r k rr r

• What is the Electric Field at r < R1? Between R1 and R2? At r > R2?

C

R1

R2

Q1

Q2

D • What happens when you connect the two spheres with a wire? (What are the

charges?)

After electrostatic equilibrium is reached, there is no charge on the inner sphere, and none on the inner surface of the shell. The charge Q1 + Q2 on the outer surface remains.

--

----- -

---

--

- - - - - --

--

Also, for r < R2 .0E

and for r > R2 r̂rQ2kE 2

1

Let’s try some numbersQ1 = 10mC R1 = 5cmQ2 = -30mC R2 = 7cm

sinner = -162 mC/m2

souter = -325 mC/m2

Electric field r < 5cm: Er(r = 4cm) = 0 N/C 5cm < r < 7cm: Er(r = 6cm) = 2.5 x 107 N/C r > 7cm: Er(r = 8cm) = -2.81 x 107 N/C

Electric field r > 7cm: Er(r = 9cm) = -2.22 x 107 N/C

B

C

D

Example 2: CylindersAn infinite line of charge passes directly through the middle of a hollow, charged, CONDUCTING infinite cylindrical shell of radius R. We will focus on a segment of the cylindrical shell of length h. The line charge has a linear charge density , and the cylindrical shell has a net surface charge density of stotal.

souter

R

sinner

h

stotal

souter

h

R

stotal

sinner

• How is the charge distributed on the cylindrical shell?• What is sinner?• What is souter?

• What is the electric field at r<R?

• What is the electric field for r>R?

A

B

C

souter

h

R

stotal

sinner

The electric field inside the cylindrical shell is zero. Hence, if we choose as our Gaussian surface a cylinder, which lies inside the cylindrical shell, we know that the net charge enclosed is zero. Therefore, there will be a surface charge density on the inside wall of the cylinder to balance out the charge along the line.• The total charge on the enclosed portion (of length h) of the line charge is:

Total line charge enclosed = h• Therefore, the charge on the inner surface of the conducting cylindrical shell is

Qinner = - h The total charge is evenly distributed along the inside surface of the cylinder. Therefore, the inner surface charge density sinner is just Qinner divided by total

area of the cylinder: sinner = - h / 2 Rh = - / 2 R • Notice that the result is independent of h.

A1 What is sinner?

souter

h

R

stotal

sinner

•We know that the net charge density on the cylinder is stotal. The charge densities on the inner and outer surfaces of the cylindrical shell have to add up to stotal. Therefore,

souter =stotal – sinner = stotal + /(2 R).

What is souter? A2

h

Gaussian surface

What is the Electric Field at r<R?

• Whenever we are dealing with electric fields created by symmetric charged surfaces, we must always first chose an appropriate Gaussian surface. In this case, for r <R, the surface surrounding the line charge is actually a cylinder of radius r.

• Using Gauss’ Law, the following equation determines the E-field: 2rhEr = qenclosed / o qenclosed is the charge on the enclosed line charge,

which is h, and (2rh) is the area of the barrel of the Gaussian surface.

The result is:

r

R

h

B

rE

r2E0

r

•As usual, we must first chose a Gaussian surface as indicated above. We also need to know the net charge enclosed in our Gaussian surface. The net charge is a sum of the following:

• Net charge enclosed on the line: h• Net charge enclosed within Gaussian surface, residing on the

cylindrical shell: Q= 2Rh stotal•Therefore, net charge enclosed is Q + h•The surface area of the barrel of the Gaussian surface is 2rh•Now we can use Gauss’ Law: 2rh E = (Q + h) / o• You have all you need to find the Electric field now.

h

R

r

Gaussian surface

stotal

What is the Electric field for r>R?C

Solve for Er to find r2rRE

00r

s

Let’s try some numbersR = 13 cm h = 168 cm stotal = 528 mC/m2 = 50mC/m

A1

A2

B • Electric Field (r<R);

Er(r = 5cm) = 1.798 x 107 N/C

C • Electric Field (r>R);

Er(r = 20cm) = 4.328 x 107 N/C

sinner = -61.21 mC/m2

souter = 589.2 mC/m2

Example 3: planesSuppose there are infinite planes positioned at x1 and x2. The plane at x1 has a positive surface charge density of s1 while the plane at x2 has negative surface charge density of s2. Find:

the x-component of the electric field at a point x>x2

the x-component of the electric field at x1<x<x2

the x-component of the electric field at a point x<x1

A

B

C

s 1 s 2

x

y

x2x1

A

Solutions

We can use superposition to find . The E-field desired is to the right of both sheets. Therefore;

0

x1 2E s1

0

2x2 2E

s0

21x 2E

ss

s 2

x2

y

x

E1E2

x1

s 1

Ex(x>x2)

E

21 EEE

B Ex(x1<x<x2)

s 1 s 2

y

xx1 x2

E2

E1

This time the point is located to the left of s2 and to the right

of s1, therefore;

0x1 2E

s10

2x2 2E

s0

21x 2E

ss

C Ex(x<x1)

When the point is located to the left of both sheets;

0x1 2E

s10

2x2 2E

s0

21x 2E

ss

s 1 s 2

y

xx1 x2

E2E1

Let’s add some numbers...

A

x1= -2m x2= 2m s1 = +2mC/m2 s2 = -3mC/m2

B

C

E1x= 1.130 x 105 N/C

E2x= -1.695 x 105 N/CEx= -0.565 x 105 N/C

E1x= 1.130 x 105 N/C

E2x= 1.695 x 105 N/CEx= 2.825 x 105 N/C

E1x= -1.130 x 105 N/C

E2x= 1.695 x 105 N/CEx= 0.565 x 105 N/C